Solid states notes class 12 IIT/JEE/NEET, Study notes of Chemistry

Download Solid states notes class 12 IIT/JEE/NEET and more Study notes Chemistry in PDF only on Docsity! Ses sio n 2 01 9-2 0 ALL EN S O L ID S TA T E E NEET SYLLABUS Solid State : Classification of solids based on different binding forces; molecular, ionic covalent and metallic solids, amorphous and crystalline solids (elementary idea), unit cell in two dimensional and three dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects, (electrical and magnetic properties, Band theory of metals, conductors, semiconductors and insulators.) (Explained in Physics) S.No. CONTENTS Page 1. Introduction 1 (a) Classification of solids 2 (b) Type of crystal systems 3 2. Analysis of cubic crystal 6 3. Crystal density 9 4. Close packing of identical solid spheres 10 5. Holes/voids in crystals 14 6. Ionic crystals 16 7. Crystal defects 20 8. Exercise -I (Conceptual Questions) 24 9. Exercise-II (Previous Years Questions) 28 10. Exercise-III (Analytical Questions) 31 11. Exercise-IV (Assertion & Reason) 32 Ses sio n 2 01 9-2 0 ALL ENOBJECTIVES After studying this unit, we will be able to : • describe general characterstics of solids. • distinction between amorphous and crystalline solids. • study crystal lattice and unit cell. • correlate the density of a substance with its unit cell properties. • explain close packing of particles. • study type of voids and ionic crystals. • describe the imperfections in solids and their effect on properties. "Chemistry without knowledge of solids ; would be a sword without a handle ; a light without brilliance ; a bell without sound" Alwin Mittasch Ses sio n 2 01 9-2 0 ALL EN Z :\ N O D E 0 2 \ B 0 A I- B 0 \ TA R G E T\ C H E M \ E N G \ M O D U LE -4 \ 1 .S O LI D S TA TE \ 0 1 -T H E O RY .P 6 5 3E Pre-Medical : Chemistry  Table (b) : The Seven Crystal Systems (Bravais Lattice) S. Name of Edge Bravais Examples No. System Lengths Angles Lattices 1. Cubic a = b = c 900 Primitive, NaCl, Zinc blende, Cu Face-centred, Body centred = 3 2. Tetragonal a= b c 900 Primitive, White tin, SnO2, TiO2, CaSO4 Body centred = 2 3. Orthorhombic a b  c 900 Primitive, Rhombic sulphur, or Rhombic Face-centred, KNO3, BaSO4 Body centred, Match Box, Duster End centred = 4 4. Rhombohedral a = b= c 900 Primitive = 1 Calcite (CaCO3), HgS (cinnabar) or Trigonal 5. Hexagonal a = b c 900, 1200 Primitive = 1 Graphite, ZnO, CdS 6. Monoclinic a b c 900900 Primitive, Monoclinic sulphur, End centred =2 Na2SO4.10H2O 7. Triclinic a b c 900 Primitive = 1 K2Cr2O7, CuSO4.5H2O, H3BO3 Total = 14 a = b = c All sides are of equal length; all angles are 900c a primitive Three angles changed body centred Trigonal (Rhombohedral) One side is changed a = b  c One side is of different length; all angles are 900 General case : All sides are different lengths; all angles are different One side length changed two angles fixed at 900 one fixed at 1200 Two side lengths made the same; one angle fixed at 1200 Length of another side is changed a  b  c Three sides are of different lengths ; all angles are 900 One angle changed Triclinic HexagonalTetragonal Orthorhombic Monoclinic c a b c a b c c a b    c a b a  c a b c a b  c a b  a = b = c All sides are of equal length;angles  Three unit cells are shown to give the hexagon Special Case a = b c  All sides are of different lengths;  face centred b primitive body centred primitive body centred face centred end centred primitive end centred primitiveprimitive primitive Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 4 E Pre-Medical : Chemistry ALLEN 1.3 Co-ordination Number The number of nearest neighbouring particles around a specific particle in a given crystalline substance is called as co-ordination number of that crystalline substance. 1.4 Packing Efficiency or Packing fraction (P.E.) "Packing efficiency is defined as the ratio of volume occupied by the atoms to the total volume of the crystalline substance" Volume occupied by atoms present in a crystal P.E. Volume of crystal = = Volume occupied by atoms present in unit cell P.E. Volume of unitcell or ´ p = 3z (4 / 3) r P.E. V Where z = number of atoms present in unit cell / Number of formula units for ionic crystals. l In a cube 1. Number of corners = 8 2. Number of faces = 6 3. Number of edges = 12 4. Number of body centre = 1 5. Number of body diagonals = 4 6. Number of face diagonals = 12 x y z a b c Face centre (6) Corner (8) Face (6) Face diagonal (12) Body diagonal (4) edge (12) Centre (1) Contribution of an atom at different lattice points of cube : l A corner of a cube is common in 8 cubes. So 1 8 th part of an atom is present at this corner of cube. l A face of a cube is common in 2 cubes. So 1 2 th part of an atom is present at the face of a cube. l An edge of a cube is common in four cubes, so 1 4 th part of the atom is present at the edge of a cube l A cube centre is not common in any another cube, so one complete atom is present at the cube centre. Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 5E Pre-Medical : ChemistryALLEN 1.5 Length of Face Diagonal and Cube Diagonal A D F B Distance between 2 adjacent face centres = =a a 2 22 Distance between 2 adjacent edge centres = =a a 2 22 Consider the triangle AFD (with the help of pyathogorous theorem) 2 2 2 2FD AF AD a a 2a= + = + = (length of face diagonal.) Consider the triangle ÐBFD (with the help of pyathogorous theorem) = +2 2BD BF FD = 2 2a ( 2a)+ = 3a (length of cube diagonal) Illustration 1. Which of the following is a non-crystalline solid ? (1) CsCl (2) NaCl (3) CaF2 (4) Glass Solution. Ans. (4) Illustration 2. Which of the following statements is incorrect about amorphous solids? (1) They are anisotropic (2) They are rigid and incompressible (3) They melt over a wide range of temperature (4) There is no orderly arrangement of particles Solution Ans. (1) Illustration 3. Assertion : Crystalline solids are anisotropic. Reason : The constituent particles are v ry closely packed. (1) A (2) B (3) C (4) D Solution. Ans. (2) 1.6 (I) CLASSIFICATION OF UNIT CELL (As per Bravais) Primitive unit cell Same type of particles are present at corners only Centered unit cell Same type of particles are present besides corners Face centered (FC) Body centered (BC) End centered (EC) Unit Cell • In end centered same type of particles are present at corners and any two opposite face centres. • End centered type of Bravais lattice is present only in orthorhombic and monoclinic type unit cell. Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 8 E Pre-Medical : Chemistry ALLEN (C) FACE CENTRED CUBIC UNIT CELL (FCC) A unit cell having lattice point at every face centre in addition to the lattice point at every corner called as face centred unit cell. i.e. in this case there are eight atoms at the eight corners of the unit cell and six atoms at the centre of six faces. The co-ordination number will be 12 and the distance between the two nearest atoms will be a 2 2 (a) Relationship between edge length 'a' and atomic radius 'r' : In FCC, along the face diagonal all atoms touches each other and the length of face diagonal is 2a . So 4r = 2a i.e. 2a r 4 = = a 2 2 a r 2 2 = (b) Number of atoms per unit cell (z) : z = 1 1 8 6 8 2 æ ö æ ö´ + ´ç ÷ ç ÷ è ø è ø = 1 + 3 = 4 atoms/unit cell Corner faces In this case one atom lies at the each corner of the cube and one atom lies at the centre of each face of the cube. It may noted that only 1 2 of each face sphere lie within the unit cell and there are six such faces. The total contribution of 8 corners is 1 8 8 æ ö´ç ÷ è ø = 1, while that of 6 face centred atoms is 1 6 2 æ ö´ç ÷ è ø = 3 in the unit cell. Hence total number of atoms per unit cell is 1 + 3 = 4 atoms. (c) Packing efficiency : æ ö ´ p ´´ p ç ÷ pè ø 3 3 3 4 a4 4z r 3 2 23P.E.= = = = 0.74 or 74% V a 3 2 [Q z =4, r= a 2 2 , V= a3] i.e. In FCC, 74% of total volume is occupied by atoms. % void space = 26 Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 9E Pre-Medical : ChemistryALLEN SUMMARY TABLE Unit cell No. of atoms Relation Co-ordination Volume occupied per unit cell between r & a Number by particles (%) Simple cube 1 8 1 8 ´ = a r 2 = 6 p ´ =100 52.4 6 Body centred 1 8 1 8 ´ + =2 a 3 r 4 = 8 p ´ = 3 100 68 8 cube (BCC) Face centred 1 1 8 6 8 2 ´ + ´ =4 a 2 r 4 = 12 p ´ =100 74 3 2 cube (FCC) Illustration 4. If the radius of an atom of an element is 75 pm and the lattice is body-centred cubic, the edge of the unit cell will be (1) 32.475 pm (2) 173.2 pm (3) 37.5 pm (4) 212.1 pm Solution. Ans. (2) Illustration 5. Assertion : A particle present at the corner of the face centred unit cell has 1/8th of its contribution of the unit cell. Reason : In any space lattice, the corner of the unit cell is always shared by the eight unit cells. (1) A (2) B (3) C (4) D Solution. Ans (3) Illustration 6. In a face centred cubic arrangement of A and B atoms where A are present at the corner and B at the face centres, A atoms are missing from 4 corners in each unit cell ? What is the simplest formula of the compound? Solution. No. of A atoms = 4 × 1 1 = 8 2 , No. of B atoms = 6 × 1 2 = 3 Formula = A1/2B3 = AB6 1.7 DENSITY OF THE CRYSTAL (d or r) If the length of edge of the unit cell is known we can calculate the density of the crystal as follow : Let length of edge of the unit cell be 'a' cm. \ Volume of the cubic unit cell = V cm3= a3 cm3 Mass of unit cell Density of the unit cell = Volume of unit cell Let mass of N particles present in a lattice = m g mass of 1 particles present in a lattice = m N g mass of z particles present in lattice = z m g N ´ ´ = ´ 3 z m d N a where z = number of particles or number of formula units (for ionic crystals) m = mass of lattice in g d = density (g/cm3) a = edge length in cm if m = M (molar mass) then N = NA ´ = ´ 3 A z M d N a Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 10 E Pre-Medical : Chemistry ALLEN • If number of particles in a lattice = N • For calculation use NA = 6 × 1023 number of SC unit cell = N number of bcc unit cell = N 2 • 1 pm = 10–12 m = 10–10 cm number of fcc unit cell = N 4 • 1 Å = 10–10 m = 10–8 cm 1. Aluminium (Molecular weight = 27) crystallises in a cubic unit cell with edge length a = 100 pm, with density, d = 180 g/cm3, then type of unit cell is (1) SC (2) BCC (3) FCC (4) HCP 2. An element has BCC unit cell, with edge length 10Å, if density is 0.2 g/cm3, then molar mass of the compound is :- (1) 240 (2) 60 (3) 35 (4) 280 3. In a tetragonal crystal :- (1) a = b =c, a = b = 90° ¹ g (2) a = b = g = 90°, a = b ¹ c (3) a = b = g = 90°, a ¹ b ¹ c (4) a = b = 90°, g=120°, a = b ¹ c 4. Edge length of a cube is 400 pm, its body diagonal would be ;- (1) 500 pm (2) 600 pm (3) 566 pm (4) 693 pm 5. A metal crystallises into two cubic phases fcc and bcc whose unit lengths are 3.5 and 3.0 Å respectively, the ratio of densities of fcc and bcc is :- (1) 1.26 (2) 1.75 (3) 2.10 (4) 1.90 1.8 CLOSE PACKING OF IDENTICAL SOLID SPHERES The solids which have non-directional bonding, their structures are determined on the basis of geometrical consideration. For such solids, it is found that the lowest energy structure is that in which each particle is surrounded by the greatest possible number of neighbours. In order to understand the structure of such solids, let us consider the particles as hard sphere of equal size in three directions. Although there are many ways to arrange the hard spheres but the one in which maximum available space is occupied will be economical which is known as closed packing now we describe the different arrangements of spherical particles of equal size. When the spheres are packed in a plane it gives two types of packing. Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 13E Pre-Medical : ChemistryALLEN Cubic close packing A C B A planes of closest packing Cubic close packing lattice formed by ABCABCA......arrangement ( type of packing) Z=4, C.N.=12, P.E.=74%. CCP is compact close packing. Hexagonal close packing :- A B A A B A B A Hexagonal close packing lattice is formed by ABABAB.....arrangement (type of packing). HCP is compact close packing. 6 fold axis Coordination number of hcp and ccp structure 9 87 4 3 X 5 6 12 11 12 10 Number of particles in hexagonal unit cell (z) = 1 1 12 2 3 1 6 6 2 ´ + ´ + ´ = C.N. = 12, P.E. = 74%, type of packing ABAB...... l Some examples of metals with their lattice types and coordination number are given in the following table. Li Na K Rb Cs Be Mg Ca Sr Ba Al Cs Y La Ti Zr Hf V Nb Ta Cr Mo W Mn Fe Tc Re Ru Os Co Rh Ir Ni Pd Pt Cu Ag Au Zn Cd Body Centre Cubic (bcc) Face Cenral Cubic Hexagonal Closed Packed (hcp) Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 14 E Pre-Medical : Chemistry ALLEN l MULTILAYER CLOSE PACKING Contents SCC BCC CCP/FCC HCP Type of packing AAAA... ABAB....... ABCABC ...... ABAB ...... packing but not packing but not close packing close packing close packing close packing No. of atoms 1 2 4 6 Co-ordination no. 6 8 12 12 Packing efficiency 52.4% 68% 74% 74% Examples IA, Ba Ca, Sr, Al Remaining Po V & Cr group Co group, Ni group, d-block elements Fe, Mn Copper group, all inert Be & Mg gases except He l In close packing, number of particles = N, number of OHV = N, Number of THV = 2N Illustration 7. The arrangement of the first two layers, one above the other, in HCP and CCP arrangements is (1) Exactly same in both cases (2) Partly same and partly different (3) Different from eath other (4) Nothing definite Solution. Ans. (1) Illustration 8. Assertion : ABAB.... pattern of close packing gives ccp arrangement. Reason : In FCC arrangement each sphere associated with two tetrahedral voids. (1) A (2) B (3) C (4) D Solution. Ans. (4) Illustration 9. “There is no difference in the arrangement of atoms in CCP and HCP structure” Do you agree with this statement ? Explain why. Solution. Statement is incorrect CCP ABC ABC ABC .... HCP AB AB AB ..... 1.9 INTERSTICES OR VOIDS OR HOLES IN CRYSTALS It has been shown that the particles are closely packed in the crystals even than there is some empty space left in between the spheres. This is known as interstices (or interstitial site or hole or empty space or voids). In three dimentional close packing (CCP & HCP) the interstices are of two types : (i) tetrahedral interstices and (ii) octahedral interstices. (A) Tetrahedral Interstices We have seen that in hexagonal close packing (HCP) and cubic close packing (CCP) each sphere of second layer touches with three spheres of first layer. Thus they, leave a small space in between which is known as tetrahedral site or interstices or the vacant space between 4 touching spheres is called as tetrahedral void. Since a sphere touches three spheres in the below layer and three spheres in the above layer hence there are two tetrahedral sites associated with one sphere. Tetrahedral void A tetrahedral interstices Tetrahedron geometry Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 15E Pre-Medical : ChemistryALLEN It may by noted that a tetrahedral site does not mean that the site is tetrahedral in geometry but it means that this site is surrounded by four spheres and by joining the centres of these four spheres forms a regular tetrahedron. In FCC, one corner and its three face centres form a tetrahedral void In FCC, two tetrahedral voids are obtained along one cube diagonal. So in FCC, 8 tetrahedral voids are present. A B C D 1 2 3 4 Te trahedra l vo id C ub e di ag on al In FCC, total number of atoms = 4 In FCC, total number of tetrahedral voids = 8 So, we can say that, in 3D close packing 2 tetrahedral voids are attached with one atom. (B) Octahedral Interstices Hexagonal close packing (hcp) and cubic close packing (ccp) also form another type of interstices which is called octahedral site.The vacant space between 6 touching spheres is called as octahedral void. In the figure, two layers of close packed spheres are shown. The spheres of first layer are shown by full circles while that of second layer by dotted circles. Two triangles are drawn by joining the centres of three touching spheres of both the layers : In fcc, 6 face centres form a octahedral void. On super imposing these triangles on one another one octahedral site is created. It may be noted that an octahedral site does not mean that the hole is octahedral in shape but it means that this site is surrounded by six nearest neighbour lattice points arranged octahedrally. (1/4 th part of octahedral void is obtained at each edge) Octahedral void (at the edge) Octahedra l vo id (a t the bo dy centre ) In FCC, total number of octahedral voids = ( 1 × 1) + (12 × 1 4 ) = 1 + 3 = 4 (Cube centre) (edge centre) In FCC, number of atoms = 4 and number of octahedral voids = 4 : number of tetrahedral voids = 8 So we can say that, in 3D close packing one octahedral void is attached with one atom. Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 18 E Pre-Medical : Chemistry ALLEN l The preferred direction of the structure with increase in the radius ratio is as follows : Plane triangular 0.225¾¾¾¾® Tetrahedral 0.414¾¾¾¾® octahedral 0.732¾¾¾¾® Cubic l LIMITING RADIUS RATIO FOR VARIOUS TYPES OF VOIDS Limiting radius Coordination Structural Example ratio = r/R Number Arrangement (Type of voids) 0.155 £ r/R < 0.225 3 Plane Trigonal Boron Oxide(B2O3) 0.225 £ r/R < 0.414 4 Tetrahedral ZnS, SiO2 0.414 £ r/R < 0.732 6 Octahedral NaCl, MgO 0.732 £ r/R < 1.000 8 Cubical CsCl Illustration 12. Assertion : In crystal lattice, the size of the cation is larger in a tetrahedral hole than in an octahedral hole. Reason : The cations occupy more space than atoms in crystal packing. (1) A (2) B (3) C (4) D Solution. Ans. (4) Illustration13. Each unit cell of NaCl consists of 14 Cl– ions and (1) 13 Na+ (2) 14 Na+ (3) 6 Na+ (4) All are wrong Solution. Ans. (1) Illustration14. A solid A+B– has NaCl type close packed structure. If the anion has a radius of 250 pm, what should be the ideal radius for the cation ? Can a cation C+ having a radius of 180 pm be slipped into the tetrahedral site of the crystal A+B– ? Give reason for your answer. Solution In Na+Cl– crystal each Na+ ion is surrounded by 6 Cl+ ions and vice versa. Thus Na+ ion is placed in octahedral hole. The limiting radius ratio for octahedral site = 0.414 or A B + - = r 0.414 R = Given that radius of anion (B–) R = 250 pm i.e. radius of cation (A+) r = 0.414 R = 0.414 × 250 pm = 103.5 pm Thus ideal radius for cation (A+) is r = 103.5 pm. We know that (r/R) for tetrahedral hole is 0.225. \ r 0.225 R = : So r = 56.25 pm Thus ideal radius for cation is 56.25 pm for tetrahedral hole. But the radius of C+ is 180 pm. It is much larger than ideal radius i.e. 56.25 pm. Therefore we can not slipped cation C+ into the tetrahedral site. Illustration15. Assertion :– In rock salt structure, the sodium ions occupy octahedral voids. Reason :– The radius ratio r+/r– in case of NaCl lies betwen 0.225 to 0.414. (1) A (2) B (3) C (4) D Solution. Ans. (3) Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 19E Pre-Medical : ChemistryALLEN T yp e of I on ic G eo m et ry C o- or di na ti on N o. o f fo rm ul a' s E xa m p le s C ry st al N um be r pe r U .C . 1 . N aC l (1 : 1 ) 6 : 6 4 N a+ + 4 C l– H al id es o f (L i, N a, K , R b) (R oc k S al t T yp e) 4 N aC l O xi de s an d su lp hi de s of (4 ) al ka lin e, e ar th m et al s ( E xc ep t B eS ) A gC l, A gB r, N H 4 X 2 . C sC l T yp e (1 : 1 ) 8 : 8 1 C s+ + 1 C l– H al id es o f 'C s' 1 C sC l T lC l, T lB r, C aS (1 ) 3 . Zn S T yp e (1 : 1 ) 4 : 4 4 Zn + 2 + 4 S –2 B eS , (Z in c B le nd e T yp e) 4 Zn S B eO , C aO , A gI , (S ph al er ite ) (4 ) C uC l, C uB r, C uI 4 . C aF 2 T yp e (1 : 2 ) 4 C a+ 2 + 8 F– B aC l 2, B aF 2 (F lu or ite T yp e) 4 C aF 2 S rC l 2, S rF 2 (4 ) C aC l 2, C aF 2 5 . N a 2O T yp e (2 : 1 ) 8 N a+ + 4 O –2 L i 2O , L i 2S (A nt ifl ou rit e) 4 N a 2O N a 2O , N a 2S (4 ) K 2O , K 2S 6 . Zn S T yp e (1 : 1 ) 4 : 4 6 Zn + 2 + 6 S –2 S am e as (W ur tz ite ) 6 Zn S sp ha le rit e an ot he r ge om et ry (6 ) of Z nS C l– C s+ 1 .1 1 S O M E I O N IC C R Y S T A L S F– C a+ 2 N a+ O –2 Zn + 2 C l– N a+ a S –2 S . N o. C s A t bo dy c en tr e of at c ub ic v oi d + ® B .C .C . C l– A t e ve ry c or ne r ® Z n+ 2 ® A t 5 0% o f T. H .V . o r a t a lte rn at e te tr ah ed ra l v oi d C .C .P . S + 2 ® E ve ry e le m en t of C .C .P . C a + 2 ® E ve ry e le m en t of C .C .P . C .C .P . F – ® A t ev er y T. H .V . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a + ® A t e ve ry O H V C .C .P . C l – ® E ve ry e le m en t o f C .C .P . N a+ ® A t ev er y T. H .V . C .C .P . O –2 ® E ve ry e le m en t of C .C .P . Z n+ 2 ® 5 0 % o f T. H .V . or (a t al te rn at e T. H .V .) H .C .P . S –2 ® E ve ry e le m en t of H .C .P . Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 20 E Pre-Medical : Chemistry ALLEN l NaCl TYPE CsCl TYPE For NaCl : For CsCl : Distance between two nearest ions ( r+ + r– ) :- Distance between two nearest ions ( r+ + r– ) :- 2r+ + 2r– = a i.e. a r r 2 + -+ = 2r+ + 2r– = 3a i.e. 3 r r a 2 + -+ = Cl— Cl— Cl— Cl— Cl— Cl— Cl— Cl— Cs+ In both structures cation and anion can exchange their position. 1.12 DEFECTS OR IMPERFECTIONS IN SOLIDS Ideal crystal : The crystal in which all the lattice points are occupied by the component particles or groups of particles is called an ideal crystal. Ideal crystals are grown up at absolute kelvin temperature (zero K). Solid state is characterised by vibratory motion about the mean position of constituent particles. At absolute zero, all the types of motions cease, and therefore crystals tend to have a perfectly ordered arrangement. As the temperature increases, molecular motions (vibratory amplitudes) increase and therefore the ions may lose their usual sites and either get missed or occupy interstitial positions in the crystal, ie., deviations from ordered arrangement take place. Any deviation from the perfectly ordered arrangement gives rise to a defect or imperfection in the crystal. Defect in crystals are produced either due to thermal effects or by adding certain impurities in the pure crystals (doping). Defects in crystals may be discussed as Crystal defects Stoichiometric defects Impurity defects (some external impurity is mixed) Non stoichiometric defects Schottky defect Frenkel defect Metal excess defect Metal deficiency defect (A) Stoichiometric Defects Stoichiometric compounds are those in which the number of positive and negative ions are exactly in the ratio as shown by their chemical formulae. Two types of defects are observed in these compounds. (a) Schottky defect, (b) Frenkel defect (a) Schottky defect : This type of defect is produced when equal number of cations and anions are missing from their respective positions leaving behind a pair of holes. The crystal as a whole remains neutral because the number of missing positive ions (cations) and negative ions (anions) is the same Ses sio n 2 01 9-2 0 ALL EN Z: \N O D E0 2\ B0 A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 1- TH EO RY .P 65 23E Pre-Medical : ChemistryALLEN Illustration 17. Assertion :– Schottky type defects is shown by crystals with high co-ordination number. Reason :– In Schottky defect equal number of cations and anions are missing from their lattice sites. (1) A (2) B (3) C (4) D Solution Ans. (2) Illustration 18. Schottky defect lowers the density of ionic crystals while Frenkel defect does not. Discuss. Solution. In Schottky defect, certain cations and anions are missing from the crystal lattice. Therefore, the density of the crystal decreases.However, in Frenkel defect the ions do not leave the lattice but they simply change their positions from lattice points to the interstitial spaces. As a result, the density of the crystal remains unchanged. 2 1. The edge length of face centred cubic unit cell having rock salt structure is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is– (1) 144 pm (2) 288 pm (3) 618 pm (4) 398 pm 2. In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the every CCP and B atoms are in OHV. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is– (1) AB2 (2) A2B (3) A4B3 (4) A3B4 3. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g ? [Atomic mass : Na = 23, Cl = 35.5] (1) 5.14 × 1021 (2) 1.28 × 1021 (3) 1.71 × 1021 (4) 2.57 × 1021 4. A solid AB has the NaCl structure. If radius of the cation A+ is 120 pm, the maximum value of the radius of the anion B– is (1) 290 pm (2) 350 pm (3) 250 pm (4) 400 pm 5. If the distnace between Na+ and Cl– ions in sodium chloride crystal is x pm, then length of the edge of the unit cell is (1) 2x pm (2) 4x pm (3) x/2 pm (4) x/4 pm 6. HCP crystal in made of A ions and B ions are arranged in 2/3rd of octahedral voids. The simplest formula of unit cell is (1) A4B6 (2) A6B4 (3) A2B3 (4) A3B2 7. A FCC crystal of O–2 (oxide) ions has M+n ions in 50% of octahedral voids, then the value of n is (1) 2 (2) 3 (3) 4 (4) 1 8. If the edge length of a KCl unit cell is 488 pm, what is the length of KCl bond if it crystallises in the FCC structure ? (1) 122 pm (2) 244 pm (3) 488 pm (4) 974 pm 9. In a solid lattice, the cation has left a lattice site and is located at interstitial position, the lattice defect is (1) interstitial defect (2) vacancy defect (3) Frenkel defect (4) Schottky defect ANSWER KEY Que. 1 2 3 4 5 Ans. 3 2 2 4 1 Que. 1 2 3 4 5 6 7 8 9 Ans. 1 4 4 1 1 4 3 2 3 BEGINNER'S BOX-1 BEGINNER'S BOX-2 Ses sio n 2 01 9-2 0 ALL EN Z: \N O DE 02 \B 0A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 2- EX ER C IS E. P6 5 24 E Pre-Medical : Chemistry ALLEN EXERCISE-I (Conceptual Questions) Build Up Your Understanding INTRODUCTION 1. a ¹ b ¹ c, a = g = 900 b ¹ 900 represents (1) tetragonal system (2) orthorhombic system (3) monoclinic system (4) triclinic system 2. The most unsymmetrical crystal system is : (1) Cubic (2) Hexagonal (3) Triclinic (4) Orthorhombic 3. Bravais lattices are of (1) 10 types (2) 8 types (3) 7 types (4) 14 types 4. The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c=0.504 nm and a=b =90° and g=120° is : (1) Cubic (2) Hexagonal (3) Orthorhombic (4) Rhombohedric ANALYSIS OF CUBIC CRYSTAL 5. In a simple cubic cell, each point on a corner is shared by (1) 2 unit cells (2) 1 unit cell (3) 8 unit cells (4) 4 unit cells 6. In a face centred cubic cell, an atom at the face contributes to the unit cell (1) 1 part (2) 1/2 part (3) 1/4 part (4) 1/8 part 7. In a body centred cubic cell, an atom at the body centre is shared by (1) 1 unit cell (2) 2 unit cell (3) 3 unit cells (4) 4 unit cells 8. Which of the following type of cubic lattice has maximum number of atoms per unit cell ? (1) Simple cubic (2) Body centred cubic (3) Face centred cubic (4) All have same 9. The number of atoms present in a unit cell of a monoatomic element of a simple cubic lattice, body- centred cubic and face centred cubic respectively are (1) 8, 9 and 14 (2) 1, 2 and 4 (3) 4, 5 and 6 (4) 2, 3 and 5 10. Which one of the following is a primitive unit cell ? (1) Simple cubic (2) BCC (3) FCC (4) bcc and fcc both 11. In a body centred cubic unit cell, a metal atom at the centre of the cell is surrounded by how many other metal atoms : (1) 8 (2) 6 (3) 12 (4) 4 12. A compound is formed by elements A and B. This crystallises in the cubic structure when atoms A are at the corners of the cube and atoms B are at the centre of the body. The simplest formula of the compound is (1) AB (2) AB2 (3) A2B (4) AB4 13. A compound formed by elements A and B has cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. The formula of the compound will be (1) A4B3 (2) A2B (3) AB3 (4) A2B3 14. Sodium metal crystallises in BCC lattice with the cell edge length (a) = 42.29 Å. What is the radius of sodium atom ? (1) 1.86 Å (2) 1.90 Å (3) 18.3 Å (4) 1.12 Å 15. An element has BCC structure having unit cells 12.08 × 1023. The number of atoms in these cells is (1) 12.08 × 1023 (2) 24.16 × 1023 (3) 48.38 × 1023 (4) 12.08 × 1022 16. A metal has BCC structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell in cm3 will be (1) 1.6 × 10–21 cm3 (2) 2.81 × 10–23 cm3 (3) 6.02 × 10—23 cm3 (4) 6.6 × 10–24 cm3 CRYSTAL DENSITY 17. An element having BCC geometry has atomic mass 50. Calculate the density of the unit cell, if its edge length is 290 pm. (1) 6.81 g cm–3 (2) 3.40 g cm–3 (3) 13.62 g cm–3 (4) 1.23 g cm–3 18. A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of metal is 5.96 g cm–3 . Find the volume of the unit cell. (1) 13.9 × 10–24 cm3 (2) 27.8 × 10–24 cm3 (3) 6.95 × 10–24 cm3 (4) 55.6 × 10–24 cm3 Ses sio n 2 01 9-2 0 ALL EN Z: \N O DE 02 \B 0A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 2- EX ER C IS E. P6 5 25E Pre-Medical : ChemistryALLEN 19. An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of the crystal is 7.2 g cm–3 , what is the atomic mass of the element? (1) 51.8 (2) 103.6 (3) 25.9 (4) 207.2 20. An element, density 6.8 g cm–3 occurs in BCC structure with cell edge 290 pm. Calculate the number of atoms present in 200 g of the element. (1) 2.4 × 1042 (2) 1.2 × 1042 (3) 1.2 × 1024 (4) 2.4 × 1024 21. An element A crystallises in fcc structure. 200 g of this element has 4.12 × 1024 atoms. The density of A is 7.2 g cm–3 Calculate the edge length of the unit cell. (1) 26.97 × 10–24 cm (2) 299.9 pm (3) 5 × 10–12 cm (4) 2.99 cm 22. Density of Li atom is 0.53 g cm–3 . The edge length of Li is 3.5 Å. Find out the number of Li atoms in an unit cell. (NA = 6.023 × 1023), (M= 6.94 g mol–1) (1) 2 (2) 8 (3) 4 (4) 6 CLOSE PACKING OF IDENTICAL SOLID SPHERES 23. The coordination number of hexagonal closest packed (HCP) structure is (1) 12 (2) 10 (3) 8 (4) 6 24. The ABAB close packing and ABC ABC close packing are respectively called as (1) hexagonal close packing(hcp) and cubic close packing (ccp) (2) ccp and hcp (3) body centred cubic (bcc) packing and hexagonal close packing (hcp) (4) hcp and bcc 25. The space occupied in BCC arrangement is (1) 74% (2) 70% (3) 68% (4) 60.4% 26. The vacant space in BCC unit cell is (1) 32% (2) 10% (3) 23% (4) 46% 27. The space occupied by spheres of equal size in three dimensions in both HCP and CCP arrangement is (1) 74% (2) 70% (3) 60.4% (4) 52.4% 28. The empty space in the HCP and CCP is about (1) 26% (2) 30% (3) 35% (4) 40% 29. Which one of the following is not a close packing ? (1) hcp (2) ccp (3) bcc (4) fcc 30. Close packing is maximum in the crystal lattice of (1) Simple cubic (2) Face centred (3) Body centred (4) Simple cubic and body centred 31. Which of the following has HCP structure ? (1) Al (2) Mg (3) Cu (4) Ni 32. All noble gases crystallise in the CCP structure except (1) Helium (2) Neon (3) Argon (4) Krypton 33. If the coordination number of an element in its crystal lattice is 8, then packing is : (1) FCC (2) HCP (3) BCC (4) None of the above HOLES / VOIDS IN CRYSTALS 34 A tetrahedral void in a crystal implies that (1) shape of the void is tetrahedral (2) molecules forming the void are tetrahedral in shape (3) the void is surrounded tetrahedrally by four spheres (4) the void is surrounded by six spheres 35. In a closest packed lattice, the number of octahedral sites as compared to tetrahedral ones will be (1) Equal (2) Half (3) Double (4) One fourth 36. The coordination number of a cation occupying an octahedral hole is (1) 4 (2) 6 (3) 8 (4) 12 37. The size of an octahedral void formed in a closest packed lattice as compared to tetrahedral void is (1) Equal (2) Smaller (3) Larger (4) Not definite 38. The coordination number of a cation occupying a tetrahedral hole is (1) 4 (2) 6 (3) 8 (4) 12 39. Number of tetrahedral voids per atom in a crystal lattice is : (1) 1 (2) 2 (3) 4 (4) 8 40. A compound contains P and Q elements. Atoms Q are in CCP arrangement while P occupy all tetrahedral sites. Formula of compound is : (1) PQ (2) PQ2 (3) P2Q (4) P3Q 41. If ‘Z’ is the number of atoms in the unit cell that represents the c losest packing sequence ABCABC---, the number of tetrahedral voids in the unit cell is equal to (1) Z (2) 2Z (3) Z/2 (4) Z/4 Ses sio n 2 01 9-2 0 ALL EN Z: \N O DE 02 \B 0A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 2- EX ER C IS E. P6 5 28 E Pre-Medical : Chemistry ALLEN AIPMT 2006 1. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs=133 and that of Br=80 amu and Avogadro number being 6.023 × 1023 mol–1 the density of CsBr is : (1) 42.5 g cm–3 (2) 0.425 g cm–3 (3) 8.25 g cm–3 (4) 4.25 g cm–3 AIIMS 2006 2. The Ca2+ and F– are located in CaF2 crystal, respectively at face centred cubic lattice points and in – (1) Tetrahedral voids (2) Half of tetrahedral voids (3) Octahedral voids (4) half of octahedral voids AIPMT 2007 3. NaCl is doped with 10–4 mol% SrCl2, the concentration of cation vacancies is – 4 A 10 Hint :Concentration of vacancies N 100 -é ù =ê ú ë û (1) 6.02 × 1015 mol–1 (2) 6.02 × 1016 mol–1 (3) 6.02 × 1017 mol–1 (4) 6.02 × 1014 mol–1 4. The fraction of total volume occupied by the atoms present in a simple cube is – (1) 6 p (2) 3 2 p (3) 4 2 p (4) 4 p AIPMT 2008 5. Percentage of free space in a body centred cubic unit is : (1) 32% (2) 34% (3) 28% (4) 30 % 6. If 'a' stands for the edge length of the cubic systems: simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these system will be respectively. (1) 1 1 a: 3 a: a 2 2 (2) 1 23 a: a: a 2 2 2 (3) 1a : a: 2 a3 (4) 1 13a: a: a 2 4 2 2 7. Which of the following statement is not correct ? (1) Molecular solids are generally volatile (2) The numbers of carbon atoms in an unit cell of diamond is 8 (3) The number of Bravais lattices in which a crystal can be categorized is 14 (4) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48 AIPMT 2009 8. Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be :- (1) 300.5 pm (2) 240.8 pm (3) 151.8 pm (4) 75.5 pm 9. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm ? (1) 108 (2) 128 (3) 157 (4) 181 AIPMT 2010 10. AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositively charged ions in the lattice is :- (1) 300 pm (2) 335 pm (3) 250 pm (4) 200 pm AIIMS 2010 11. In a trigonal crystal, which statement is incorrect (1) All the axial length and axial angle are equal (2) All the three axial lengths are equal (3) All the three axial angles are equal (4) Two axial angles are same and one is different AIPMT Mains. 2011 12. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm the radius of the anion (Y–) will be :- (1) 165.7 pm (2) 275.1 pm (3) 322.5 pm (4) 241.5 pm EXERCISE-II (Previous Year Questions) AIPMT/NEET & AIIMS (2006-2018) Ses sio n 2 01 9-2 0 ALL EN Z: \N O DE 02 \B 0A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 2- EX ER C IS E. P6 5 29E Pre-Medical : ChemistryALLEN AIIMS 2011 13. Schottky defect is– (1) vacancy of ions (2) Delacolization of ions (3) Interstitial vacancy of ions (4) vacancy of only cation AIPMT Pre. 2012 14. The number of octahedral void(s) per atom present in a cubic-close-packed structure is: (1) 2 (2) 4 (3) 1 (4) 3 15. A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is: (1) 144 pm (2) 204 pm (3) 288 pm (4) 408 pm AIPMT Mains 2012 16. Structure of a mixed oxide is cubic close-packed (CCP). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is: (1) A2B3O4 (2) AB2O2 (3) ABO2 (4) A2BO2 AIIMS 2012 17. In BCC structure, contribution of corner and central atom is :- (1) 1 ,1 8 (2) 1 1 , 4 8 (3) 1 1 , 8 2 (4) 1 1, 2 NEET-UG 2013 18. The number of carbon atoms per unit cell of diamond unit cell is :- (1) 1 (2) 4 (3) 8 (4) 6 19. A metal has a FCC lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72g cm–3. The molar mass of the metal is :- (1) 20g mol–1 (2) 40g mol–1 (3) 30g mol–1 (4) 27g mol-1 AIIMS 2013 20. In a solid, atom M occupy CCP lattice and 1/3rd of tetrahedral voids are occupied by atom N. Find formula of solid formed by M and N :- (1) M3N2 (2) M2N3 (3) M4N3 (4) M3N4 AIPMT 2014 21. If a is the length of the side of a cube, the distance between the body centered atom and one corner atom in the cube will be : (1) 2 a 3 (2) 4 a 3 (3) 3 a 4 (4) 3 a 2 AIIMS 2014 22. In any compound, atom B occupies in HCP packing and A is 1/3rd of tetrahedral voids. Then emperical formula of compound is :- (1) B2A3 (2) B3A2 (3) AB (4) AB2 AIPMT 2015 23. A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom? (1) 127 pm (2) 80 pm (3) 108 pm (4) 40 pm Re-AIPMT 2015 24. The vacant space in BCC lattice unit cell is : (1) 23% (2) 32% (3) 26% (4) 48% 25. The correct statement regarding defects in crystalline solids is :- (1) Frenkel defect is a dislocation defect (2) Frenkel defect is found in halides of alkaline metals (3) Schottky defects have no effect on the density of crystalline solids (4) Frenkel defects decrease the density of crystalline solids AIIMS 2015 26. Defects in NaCl & AgCl crystals are respectively (1) Schottky, Frenkel (2) Frenkel, Schottky (3) Both Schottky (4) Both Frenkel Ses sio n 2 01 9-2 0 ALL EN Z: \N O DE 02 \B 0A I-B 0\ TA RG ET \C H EM \E N G \M O D U LE -4 \1 .S O LID S TA TE \0 2- EX ER C IS E. P6 5 30 E Pre-Medical : Chemistry ALLEN EXERCISE-II (Previous Year Questions) ANSWER KEY Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. 4 1 3 1 1 4 4 3 2 2 4 4 1 3 3 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. 2 1 3 4 1 4 2 1 2 1 1 1 2 1 1 Que. 31 32 33 34 35 36 37 38 39 40 Ans. 1 1 3,4 4 3 3 1 2 1 2 27. What will be value of Cs Cl r r+ -+ in CsCl crystal having edge length 'a' (1) a 3 2 (2) a 3 (3) a 3 4 (4) a 2 2 NEET-I 2016 28. Lithium has a BCC structure. Its density is 530 kg m–3 and its atomic mass is 6.94 g mol–1. Calculate the edge length of a unit cell of Lithium metal. (NA = 6.02 × 1023 mol–1) (1) 154 pm (2) 352 pm (3) 527 pm (4) 264 pm 29. The ionic radii of A+ and B– ions are 0.98 × 10–10m and 1.81 × 10–10 m. The coordination number of each ion in AB is :- (1) 6 (2) 4 (3) 8 (4) 2 NEET-II 2016 30. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+) and fluoride ion (F–) are (1) 8 and 4 (2) 4 and 8 (3) 4 and 2 (4) 6 and 6 AIIMS 2016 31. In a crystalline solid, X atoms occupy hcp and 1 4 th of tetrahedral voids are occupied by Y atoms. Then empirical formula of crystalline solid is :- (1) X2Y (2) XY2 (3) X4Y (4) XY 32. A metal crystallizes in bcc lattice having a radius 4Å. Then what will be edge length of its unit cell? (1) 16 Å 3 (2) 16 3Å (3) 16 Å 2 (4) 16 2Å NEET(UG) 2017 33. Which is the incorrect statement ? (1) Density decreases in case of crystals with Schottky's defect (2) NaCl(s) is insulator, silicon is semiconductor, silver is conductor, quartz is piezo electric crystal (3) Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal (4) FeO0.98 has non stoichiometric metal deficiency defect AIIMS 2017 34. Order of packing efficiency is :- (1) HCP > fcc > BCC > simple cubic (2) HCP » FCC > simple cubic > BCC (3) BCC > HCP > simple cubic > fcc (4) HCP » FCC > BCC > simple cubic 35. In a cube X atoms are present at corners and Y atoms are at face centres of a cube. The number of X and Y are :- (1) 8, 8 (2) 6, 8 (3) 8, 6 (4) 6, 6 NEET-II 2018 36. Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is (1) 3 2 (2) 4 3 3 2 (3) 3 3 4 2 (4) 1 2 AIIMS 2018 37. F-centre is produced due to :- (1) Unpaired electron occupied anionic vacancies (2) Monovalent ion is replaced by bivalent ion (3) Ions present in cationic vacancies (4) Disslocation of ions 38. Which of the following is correct for ZnO :- (1) Impart color due to charge transfer (2) On heating it turns yellow (3) Zn can reduce hydrogen at room remprature (4) It sublimises on heating 39. Among the following the molecular solid is :- (1) SO2(solid) (2) SiC (3) Graphite (4) ZnS 40. In any compound, atoms of A occupies in HCP packing and atoms of B is at 1 3 rd of tetrahedral voids. Then emperical formula of compound is :- (1) A2B3 (2) A3B2 (3) B3A (4) BA2